Question

Explore the principal value of the function sin inverse using a unit curcle with the an appropriate figure of the inverse function​

Answer 1

Step-by-step explanation:

The principal value of sin−1−1 x for x > 0, is the length of the arc of a unit circle centred at the origin which subtends an angle at the centre whose sine is x. For this reason sin^-1 x is also denoted by arc sin x. Similarly, cos−1−1 x, tan−1−1  x, csc−1−1  x, sec−1−1  x and cot−1−1 x are denoted by arc cos x, arc tan x, arc csc x, arc sec x.

1. Find the principal values of sin−1−1 (- 1/2)      

Solution: 

If θ be the principal value of sin−1−1 x then - π2π2 ≤ θ ≤ π2π2.

Therefore, If the principal value of sin−1−1 (- 1/2) be θ then sin−1−1 (- 1/2) = θ

⇒ sin θ = - 1/2 = sin (-π6π6) [Since, - π2π2 ≤ θ ≤ π2π2]

Therefore, the principal value of sin−1−1 (- 1/2) is (-π6π6).

2. Find the principal values of the inverse circular function cos−1−1 (- √3/2)

Solution:

 If the principal value of cos−1−1 x is θ then we know, 0 ≤ θ ≤ π.

Therefore, If the principal value of cos−1

Answer 2

Answer:

the principal value of $sinx^{-1}$₋$1 (- 1/2)$ is $(-\pi6\pi6)$.

(Trigonometric function solutions with a value between 0 and 2 are known as principal values of trigonometric functions. The primary values of trigonometric functions are values smaller than 2, which is the interval at which the trigonometric function's value repeats.)

Step-by-step explanation:

• The principal value of $sinx^{-1}$ x for $x > 0$, is the length of the arc of a unit circle centred at the origin which subtends an angle at the centre whose sine is $x$. For this reason $sin^{1} x$ is also denoted by arc $sin x$. Similarly, $cos^{1}$₋$1 x$, $tan^{-1} -1 x,$ $csc^{-1} -1 x$, $sec^{-1} -1 x$ and $cot^{-1} -1 x$ are denoted by arc cos x, arc tan x, arc csc x, arc sec x.

• If θ be the principal value of $sin^{-1} -1 x$ then $- \pi2\pi 2 \leq$Θ$\leq \pi 2\pi 2$.

• Therefore, If the principal value of $sin^{-1}-1 (- 1/2)$ be Θ then $sin^{-1} -1 (- 1/2) =$Θ
  ⇒ $sin θ = - 1/2 = sin (-\pi 6\pi 6)$ [Since, $- \pi 2\pi 2 \leq$Θ$\leq \pi 2\pi 2$ ]
  Therefore, the principal value of $sin^{-1} -1 (- 1/2)$ is $(-\pi 6\pi 6)$.

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